Math Is Fun Forum

No need.  Look on the left for something that is also on the right.  If found cancel them to make a simpler equation.

eg k squared will go.

Bob

He sure did. So I think we'll take that as correct!

Now on to question 2.

As you don't yet know 'n' you cannot just write out the rows of Pascal's triangle.  And your calc. won't be able to either because it'll want to know the value of n.

Luckily, there's a way of calculating the numbers for any row without having to know the ones before.

For the nth row it is:

The third and fourth terms both have coefficient B so we can put them equal

Most of that cancels out so over to you again.  Simplify as much as you can.  What are you left with?

Bob

Great.

So for power 6 you want

1     6     15      20       15      6      1

and then the terms are

I stopped at x squared because the question does too.  (just corrected an error there. whoops)

Ok so far?

Bob

hi Bry,

Good to 'hear' from you.  This is the 'chat'.

Is your PE hw to roll about on the floor laughing?

Let's start with question 1.  Have you understood how to do it?

Bob

http://en.wikipedia.org/wiki/Manifold

Bob

hi aqa,

You seem to have been much more careful than me in considering all possible cases.  I like all your suggestions.

Do you think I can still claim this as a total success for my teaching, as the 'pupil' has overtaken the 'teacher' ?

Bob

hi aqa

It has got to be 40 years since I met reduced row echelon matrices (rrem) in my degree and I haven’t used them since,  so I’m a bit rusty. 
That’s why (and it was late at night for me) I pointed you to Wiki.  This site does get some criticism for lack of accuracy but I’ve always found it good for mathematical topics.

Now I’ve had a chance to think it through, I’m ready to give you what is, I hope, a better answer.

The only use I’ve come across for rrem is a way of solving simultaneous linear equations.  (That is equations where there are no square, cube, or higher powers)

The procedure is no quicker or easier than other methods but it has the advantage that it is completely prescriptive ... you do exactly the same thing for every set of equations.  So it is ideal if you want to write a computer program to solve equations.  The Wiki page gives some pseudocode for this if you are interested and have the time to write and debug the program.

I’ll demonstrate the procedure by way of an example.  To use the procedure you must have all the equations in the form

Suppose the equations are

Step 1.  Build a 3 by 4 matrix out of these numbers

Now to change into rrem

Step2. Divide the top row by its first element (2)

Step 3.  Make new row 2 := 3 x row 1 - old row 2.   Make new row 3 := row `1 - old row 3

later edit:  Looking this over, I think this step may be back to front.  The 'rust' had not quite cleared at this stage.  But it works out ok so it doesn't matter a lot unless you are trying to build a computer procedure.  If so, the later steps show the correct way to create the ones and zeros correctly.

This puts a one and two zeros into the first column.

Step 4.

Divide the second row by the second element in the row.

Step 5.  Make new row 1 := old row 1 + 0.5 x row 2.   Make new row 3 := old row 3 + 1.5 x row 2.

That has made the second column into a zero, a one and another zero.

Step 6.  Divide row 3 by the third element in the row.

Step 7.  Make new row 1 := old row 1 + 3/7 of row 3.   Make new row 2 := old row 2 - 1/7 of row 3

Now you can translate that back into three equations

This procedure will always do that.  If there are more equations than unknowns something funny will show whether the 'excess' equations are consistent with the ones you use to solve the equations.  If you have insufficient equations then the procedure will not get completed.

Have you sorted out your original query?  I have an answer if you are stuck. 

Bob

hi aqa

I think it means start with an entirely new 2 x 3 matrix in rref.  How many of these can you find?  (From the form of the question you put * for anything that is not a zero or one and count all such as one possibility.)

Bob

hi aqa,

You are welcome.  Post again, if you need help.

Bob

Strictly (read the title of this thread)

Bob

ps  Am working on some new Smilies

pps gAr: Sorry I feel guilty now.  I asked you to remove the answer and I've gone and plastered it back on.  I'll work on a "Guilty look" Smily first.  Meanwhile best fit =

Put your answer back and then it shows you have prior claim.

I'm adding it to my ever growing list of problems I'll come back to.

Bye for now,

Bob 

hi bobbym and gAr

How do you justify it?

Well, it is 'symmetrical' in a, b and c (by which I mean, if you swap all instances of, say, a with those of, say, b, the equation is unchanged.

So the 'answer' must also be symmetrical; ie.  whatever the 'a' bit looks like, the 'b' and 'c' bits will look the same.  So that answer is an intelligent guess.

gAr: I wasn't putting up the answer just yet because I was waiting for the OP to come back.  Otherwise, why are we spending time on his problem? (except obviously, it's lovely to chat to you all).  As he is off-line, you could edit it out now we've seen it.

Cannot find a smily for 'being conspiratorial'

Bob